Section 3.3: Integrals in Work
When forces vary with position, integrals are the most powerful tool for calculating work. The work done is represented as the area under a force–displacement curve, which can often be expressed using definite integrals.
\( W = \int_{x_i}^{x_f} F(x) \, dx \)
For forces that depend on other variables (like angle or time), similar integrals apply:
- Rotational work: \( W = \int_{\theta_i}^{\theta_f} \tau(\theta) \, d\theta \)
- Work in multiple dimensions: \( W = \int \vec{F} \cdot d\vec{r} \)
Example 1
A particle experiences force \( F(x) = 3x^2 \). Find the work done as the particle moves from \( x=1 \) m to \( x=3 \) m.
\( W = \int_1^3 3x^2 \, dx = [x^3]_1^3 \)
\( = 27 - 1 = 26 \, \text{J} \)
Example 2
A torque \( \tau(\theta) = 5\theta \) (N·m) acts on a rotating wheel. Find the work done as it rotates from \( \theta=0 \) to \( \theta=\pi \) radians.
\( W = \int_0^{\pi} 5\theta \, d\theta = \tfrac{5}{2}\theta^2 \big|_0^{\pi} \)
\( = \tfrac{5}{2}\pi^2 \, \text{J} \)
Practice Problems
- Find the work done if \( F(x) = 4x \) acts from \( x=0 \) to \( x=5 \) m.
- Compute the work when \( F(x) = 2x^3 \), \( x=1 \) to \( x=2 \) m.
- A torque \( \tau(\theta) = 3\sin\theta \). Find work for \( \theta=0 \) to \( \theta=\pi \).
- Evaluate \( W = \int_0^2 (6-2x)\,dx \).
- A force field \( \vec{F} = (2x)\hat{i} + (3y)\hat{j} \). Compute work along a straight line from (0,0) to (1,1).